I am asked to solve the following ODE involving constants $\alpha, L, V_0 > 0$ and $E < 0$:
$$-\psi'' - \alpha V_0\psi\cdot [\delta(x)+\delta(x-L)] = \alpha E\psi.$$
In particular, we want solutions $\psi:\mathbb{R}\to \mathbb{C}$ that are:
- Continuous
- $L^2$
We are given that the solutions look something like this (shifted):
Now pretend we didn't know this. Here is my attempt at arriving at these functions:
Away from $0$ and $L$, the ODE is just the wave equation, whose solutions are always of the form $a_1e^{-i\sqrt{\alpha E}x}+a_2e^{i\sqrt{\alpha E}x}.$ So we know $\psi$ is also of this form piece-wise: $$\psi(x)=\left\lbrace \begin{array}{ll} A_1e^{\sqrt{|\alpha E|}x}+A_2e^{-\sqrt{|\alpha E|}x} & x<0 \\ B_1e^{\sqrt{|\alpha E|}x}+B_2e^{-\sqrt{|\alpha E|}x} & 0\leq x< L \\ C_1e^{\sqrt{|\alpha E|}x}+C_2e^{-\sqrt{|\alpha E|}x} & x\geq L \\ \end{array} \right.$$ (I expanded a little here, since $E<0\Rightarrow \sqrt{\alpha E}=i\sqrt{|\alpha E|}$)
- For $\psi$ to be integrable, we also need the exponentials with infinite > support be integrable. Then $A_2=C_1=0.$
Then we can simplify it: $$\psi(x)=\left\lbrace \begin{array}{ll} A_1e^{\sqrt{|\alpha E|}x} & x<0 \\ B_1e^{\sqrt{|\alpha E|}x}+B_2e^{-\sqrt{|\alpha E|}x} & 0\leq x< L \\ C_2e^{-\sqrt{|\alpha E|}x}& x\geq L \\ \end{array} \right.$$ All that's left to do is to find the space of coefficients $A_i,B_i,C_i$ so that it is continuous and satisfies the DE.
- For $\psi$ to be continuous, $\lim_{x\to 0^+}\psi(x)=\lim_{x\to 0^-}\psi(x)$ and likewise at $L$. So $A_1=B_1+B_2,$ and a similar equation at $L.$
- To satisfty the ODE at $0$ and $L$, because of the deltas, we know that the derivative must have a discintinuity of size $-\alpha V_0 \psi$ there. In particular: $(\lim_{x\to 0^+}\psi'(x))-(\lim_{x\to 0^-}\psi'(x))=-\alpha V_0 \psi(0),$ so computing these terms, $(\sqrt{|\alpha E|}B_1 - \sqrt{|\alpha E|}B_2)- (\sqrt{|\alpha E|}A_1) = -\alpha V_0A_1$. We have a similar equation at $L$.
Now we have 4 equations relating all the coefficients to each other. I solved this system and plotted the solutions. The problem is, varying values of $\alpha, L, V_0, E$, I can only get solutions that look like the 'even' graph.
Where did the odd solutions disappear in this analysis and how do you get them back?
As stated, the solution to the ODE is $$ \psi(x) = \begin{cases} A_1 e^{\sqrt{-\alpha E} x} & x < 0,\\ B_1 e^{\sqrt{-\alpha E} x} + B_2 e^{-\sqrt{-\alpha E} x} & 0 < x < L, \\ C_2 e^{-\sqrt{-\alpha E} x} & L < x.\\ \end{cases} $$ Writing the jump and continuity conditions in matrix form, we have $$ \begin{pmatrix} 1 & -1 & -1 & 0\\ 0 & e^{2 L \sqrt{-\alpha E}} & 1 & -1\\ 1-\sqrt{-\frac{\alpha V_0^2}{E}} & -1 & 1 & 0\\ 0 & e^{2 L \sqrt{-\alpha E}} & -1 & 1-\sqrt{-\frac{\alpha V_0^2}{E}} \end{pmatrix} \begin{pmatrix}A_1 \\ B_1 \\ B_2 \\ C_2\end{pmatrix} = 0, $$ so either you have the trivial solution or the determinant of the matrix is zero. Hence, $$ f(E) = e^{2 L\sqrt{-\alpha E}}\left(1 - \frac{4 E}{\alpha V_0^2} - 4\sqrt{-\frac{E}{\alpha V_0^2}}\right) = 1. $$ This is an interesting equation. It's clear that is trivially satisfied if $E = 0$, but this leads to a nonphysical solution.
The function $f$ is nonnegative, has a unique double root at $E = -\frac{\alpha V_0^2}{4}$ and grows as $c_1|E|e^{c_2\sqrt{|E|}}$. This means that it exists at least one solution of the transcendental equation such that $f(E_*) = 1$. Moreover,
$$-\alpha V_0^2 < E_* < -\frac{\alpha V_0^2}{4}.$$
Also, if $\alpha V_0 L > 2$, then $$\lim_{E\to 0^-}f'(E) = -\infty,$$ wich implies that $f$ has a local maximum $E_M$ such that $f(E_M) > 1$, i.e., $$E_M = -\tfrac{(\alpha V_0 L - 2)^2}{4 \alpha L^2}.$$ As consequence, there is a second solution such that $f(E^*) = 1$, where $$-\frac{\alpha V_0^2}{4} < E^* < -\tfrac{(\alpha V_0 L - 2)^2}{4 \alpha L^2}.$$
Combining this results, we have:
For $L=1$, $\alpha = 1$, $V_0 = \frac{3}{2}$, this is how the eigenfunction looks like:
For $L=1$, $\alpha = 1$, $V_0 = \frac{5}{2}$, this are the two eigenfunctions:
Nice!